Question

Explain why it is always possible to express any homogeneous differential equation in the form$\mathit{M}\left(x,y\right)\mathit{d}\mathit{x}\mathbf{=}\mathit{N}\left(x,y\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

You might start by proving that

$\mathit{M}\left(x,y\right)\mathbf{=}{\mathit{x}}^{\mathbf{\alpha }}\mathit{M}\left(1,\frac{y}{x}\right)\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{N}\left(x,y\right)\mathbf{=}{\mathit{x}}^{\mathbf{\alpha }}\mathit{N}\left(1,\frac{y}{x}\right)$

Short answer

As a result, we can determine the function $F\left(\frac{y}{x}\right)$as follows: $F\left(\frac{y}{x}\right)=-\frac{M\left(1,{\displaystyle \frac{y}{x}}\right)}{N\left(1,{\displaystyle \frac{y}{x}}\right)}$

Step-by-step solution

Homogeneous Differential equation

A homogeneous differential equation is one that can be expressed in the form

$\frac{dy}{dx}=F\left(\frac{y}{x}\right)$

From the change of variable, this equation can be turned into a separable variable equation:$y=ux\to dy=dux+udx$

Given the homogeneous differential equation$M\left(x,y\right)dx+N\left(x,y\right)dy=0$where the functions$M\left(x,y\right),N\left(x,y\right)$satisfy the condition:

$M\left(tx,ty\right)=t^{\alpha }M\left(x,y\right),N\left(tx,ty\right)=t^{\alpha }N\left(x,y\right)$using the value $tx=1\to t=\frac{1}{x}$We can rewrite the above expression as follows

Solve the expression for M and N

$$\begin{gathered} M\left(1,\frac{y}{x}\right)={\left(\frac{1}{x}\right)}^{\alpha }M\left(x,y\right) \\ M\left(1,\frac{y}{x}\right)=\left(\frac{1}{x^{\alpha }}\right)M\left(x,y\right) \\ M\left(x,y\right)=x^{\alpha }M\left(1,\frac{y}{x}\right) \\ N\left(1,\frac{y}{x}\right)={\left(\frac{1}{x}\right)}^{\alpha }N\left(x,y\right) \\ N\left(1,\frac{y}{x}\right)=\left(\frac{1}{x^{\alpha }}\right)N\left(x,y\right) \\ N\left(x,y\right)=x^{\alpha }N\left(1,\frac{y}{x}\right) \end{gathered}$$

Rewrite the differential equation

$$\begin{gathered} M\left(x,y\right)dx+N\left(x,y\right)dy=0 \\ {x}^{\alpha }M\left(1,\frac{y}{x}\right)dx+x^{\alpha }N\left(1,\frac{y}{x}\right)dy=0 \\ {x}^{\alpha }N\left(1,\frac{y}{x}\right)dy=-x^{\alpha }M\left(1,\frac{y}{x}\right)dx \\ \frac{dy}{dx}=-\frac{x^{\alpha }M\left(1,\frac{y}{x}\right)}{x^{\alpha }N\left(1,\frac{y}{x}\right)} \\ \frac{dy}{dx}=-\frac{M\left(1,\frac{y}{x}\right)}{N\left(1,\frac{y}{x}\right)} \end{gathered}$$

As a result, we can determine the function $F\left(\frac{y}{x}\right)$as follows: role="math" localid="1663951397133" width="140" height="69">Fyx=-M1,yxN1,yx